Chapter 5.2 Sequences

In this section, we develop the principle of finite approximation through sequences. We will define what a sequence is, how to describe sequences, and discuss properties certain sequences posses.

Analytic Properties of the Real Numbers

To discuss closeness, we need to define what a sequence is.

Sequence

For any set \(X\), a sequence \(a\) in \(X\) is a function with either domain \(\mathbb{N}\) and codomain \(X\) (known as an infinite sequence), or ,for some natural number \(N\), domain of \(\{1,2,\dots, N\}\) and codimain \(X\) (known as a finite sequence).

Use the figure below to understand what a sequence is.

In other words, a sequence is a function with a domain equal to \(\mathbb{N}\) or a finite subset of \(\mathbb{N}\). This is different from the functions we have seen before which usually have a domain consistning of some union of intervals.

For an infinite sequence \(a\), we will denote it by \((a_n)\) and for each \(n\) in \(\mathbb{N}\), \[a(n)=a_n,\] where we read the above as “a sub n”. We also sometimes view \((a_n)\) as an “infinite” tuple \((a_1,a_2,a_3,\dots,a_k,\dots)\) where \(a_k\) refers to the \(k^{\text{th}}\) term of \(a\).

When \((a_n)\) is a finite sequence with domain \(\{1,\dots,N\}\), we will denote it as \((a_n)_{n\in \{1,\dots,N\}}.\)

In the next example, practice writing out the terms in the finite sequence.

Example 1

Express \((a_n)_{n\in\{1,\dots,6\}}\) as a \(6\)-tuple, where for all \(n\) in \(\{1,\dots,6\}\), \[a_n=3n-2.\]

To determine \(a_1,a_2,a_3,a_4,a_5\) and \(a_6\), evaluate \(a_n\) at the indicated subscript:

  • \(a_1=3\cdot 1-2=1\)
  • \(a_2=3\cdot 2-2=4\)
  • \(a_3=3\cdot 3-2=7\)
  • \(a_4=3\cdot 4-2=10\)
  • \(a_5=3\cdot 5-2=13\)
  • \(a_6=3\cdot 6-2=16.\)

The \(6\)-tuple is \[(1,4,7,10,13,16).\]

For next example, practice working with infinite sequences.

Example 2

Write the first four terms of \((a_n)\), where for each natural number \(n\), \(a_n\) is given by:

  1. \(a_n = 2n-1\);

  2. \(a_n = n^2\);

  3. \(a_n = \tfrac{1}{3n}\)

  1. The first four terms:
    • \(a_1=2(1)-1=2-1=1\)
    • \(a_2=2(2)-1=4-1=3\)
    • \(a_3=2(3)-1=6-1=5\)
    • \(a_4=2(4)-1=8-1=7\)
  2. The first four terms:
    • \(a_1=(1)^2=1\)
    • \(a_2=(2)^2=4\)
    • \(a_3=(3)^2=9\)
    • \(a_4=(4)^2=16\)
  3. The first four terms:
    • \(a_1=\frac{1}{3\cdot 1}=\frac{1}{3}\)
    • \(a_2=\frac{1}{3\cdot 2}=\frac{1}{6}\)
    • \(a_3=\frac{1}{3\cdot 3}=\frac{1}{9}\)
    • \(a_4=\frac{1}{3\cdot 4}=\frac{1}{12}\)

Because sequences are functions, we can talk about what it means for a sequence to be increasing, strictly increasing, decreasing, and strictly decreasing.

Increasing, Strictly Increasing, Decreasing, Strictly Decreasing and Monotone Sequence

Take \((a_n)\) to be a sequence.

  • The sequence \((a_n)\) is increasing if for all natural numbers \(n\), \[a_{n+1}\geq a_n.\]
  • The sequence \((a_n)\) is strictly increasing if for all natural numbers \(n\), \[a_{n+1}> a_n.\]
  • The sequence \((a_n)\) is decreasing if for all natural numbers \(n\), \[a_{n+1}\leq a_n.\]
  • The sequence \((a_n)\) is strictly decreasing if for all natural numbers \(n\), \[a_{n+1}< a_n.\]

A sequence \((a_n)\) is (stritcly) monotone if it is either (strictly) increasing or (strictly) decreasing.

Use the pictures below to understand the definition.

An equivalent condition to showing a sequence is increasing is to show \[a_{n+1}-a_n\geq 0 \quad\text{rather than}\quad a_{n+1}\geq a_n.\]

Similarly, to show a sequence is decreasing, show instead \[a_{n+1}-a_n\leq 0 \quad\text{rather than}\quad a_{n+1}\leq a_n.\]

In this next example, show that the sequence is either increasing or decreasing. Pay attention to how show the given sequence is increasing or decreasing.

Example 3

Show that the sequence \((a_n)\) is increasing and the sequence \((b_n)\) is decreasing, where

  1. \(a_n=\frac{n}{n+1}\)
  2. \(b_n=\frac{n+2}{n+1}\).

  1. The \(n\) and \(n+1\) term are \(a_n=\tfrac{n}{n+1}\) and \(a_{n+1}=\tfrac{n+1}{n+2}\). To conclude \((a_n)\) is increasing, show for all \(n\) that \(a_{n+1}>a_n\). This is equivalent to showing that \(a_{n+1}-a_n>0\). Rewrite the expression \(a_{n+1}-a_n\) like this: \[\begin{align*} a_{n+1}-a_n&=\tfrac{n+1}{n+2}-\tfrac{n}{n+1}\\ &=\tfrac{(n+1)(n+1)}{(n+2)(n+1)}-\tfrac{n(n+2)}{(n+2)(n+1)}\\ &=\tfrac{n^2+2n+1}{(n+2)(n+1)}-\tfrac{n^2+2n}{(n+2)(n+1)}\\ &=\tfrac{n^2+2n+1-n^2-2n}{(n+2)(n+1)}\\ &=\tfrac{1}{(n+2)(n+1)}. \end{align*}\] The expression \(a_{n+1}-a_n=\tfrac{1}{(n+2)(n+1)},\) is a quotient of two expressions that are always positive for all \(n\in\mathbb{N}\). Thus \(a_n\) is increasing.

  2. The \(n\) and \(n+1\) term are \(b_n=\tfrac{n+2}{n+1}\) and \(b_{n+1}=\tfrac{n+3}{n+2}\). To conclude \((b_n)\) is decreasing, show for all \(n\) that \(b_{n+1}>b_{n}\). This is equivalent to showing that \(b_{n+1}-b_{n}>0.\) Rewrite the expression \(b_{n+1}-b_n\) like this:

\[ \begin{align*} b_{n+1}-b_{n}&=\tfrac{n+3}{n+2}-\tfrac{n+2}{n+1}\\ &=\tfrac{(n+1)(n+3)}{(n+2)(n+1)}-\tfrac{(n+2)(n+2)}{(n+2)(n+1)}\\ &=\tfrac{n^2+4n+3}{(n+2)(n+1)}-\tfrac{n^2+4n+4}{(n+2)(n+1)}\\ &=-\tfrac{1}{(n+2)(n+1)}. \end{align*} \] The expression \(b_{n+1}-b_{n}=-\tfrac{1}{(n+2)(n+1)}\) is a quotient of two expressions that are always positive for all \(n\in\mathbb{N}\), but he minus sign makes the overall expression negative. Thus \(b_n\) is decreasing.

Another kind of sequence is a sequence of intervals in \(\mathbb{R}\). Specifically,it is a squence with codomain equal to the set of intervals in \(\mathbb{R}.\)

For any finite sequence of interval \((I_n)_{n\in\{1,\dots,N\}},\)

  • the union \(\displaystyle\bigcup_{n\in\{1,\dots,N\}}I_n\) is a set of real numbers with the property that a real number \(x\) is in this set if and only if there is an \(n\) in \(\{1,\dots,N\}\) so that \(x\) is in \(I_n\).
  • the intersection \(\displaystyle\bigcap_{n\in\{1,\dots,N\}}I_n\) is a set of real numbers with the property that a real number \(x\) is in this set if and only if for all \(n\) in \(\{1,\dots,N\}\), \(x\) is in \(I_n\).

Practice computing the union and intersection of a sequence of intervals with this next example. It may be helpful to draw the intervals out and then take the appropriate union or intersection of them.

Example 4

Take \((I_n)_{n\in\{1,2,3,4\}}\) to be the finite sequence of intervals that is given by \[I_1=[-2,8), I_2=(-3,6),I_3=[1,12],\text{ and }I_4=(0,4).\]

  1. Express as an interval the union \(\displaystyle\bigcup_{n\in\{1,2,3,4\}}I_n\).

  2. Express as an interval the intersection \(\displaystyle\bigcap_{n\in\{1,2,3,4\}}I_n\).

  1. We have that \[\displaystyle\bigcup_{n\in\{1,2,3,4\}}I_n=(-3,12]\]

  2. We have that \[\displaystyle\bigcap_{n\in\{1,2,3,4\}}I_n=[1,4).\]

When we have an infinite sequence of intervals \((I_n)\),

  • the union \(\displaystyle\bigcup_{n\in\mathbb{N}}I_n\) is a set of real numbers with the property that a real number \(x\) is in this set if and only if there is an \(n\) in \(\mathbb{N}\) so that \(x\) is in \(I_n\).
  • the intersection \(\displaystyle\bigcap_{n\in\mathbb{N}}I_n\) is a set of real numbers with the property that a real number \(x\) is in this set if and only if for all \(n\) in \(\mathbb{N}\), \(x\) is in \(I_n\).

Working with unions and intersections in the setting of infinite sequences is much more subtle than working with unions and intersections in the setting of finite sequences.

To understand why this is the case, let’s first discuss length. A rational number can be expressed as a quotient of an integer and a natural number. Any computations involving rational numbers involves only arithmetic. Lengths that are rational numbers are easy to work with, but not all lengths are rational.

For example, the hypothenus of a right triangle with both edge lengths equal to \(1\) has a hypotenuse of length \(c\), where \[c^2=2.\]

There is no rational number \(c\) with this property. And so not all lengths are rational numbers. This is why we need real numbers.

Although there are more real numbers than rational numbers, the idea of magnitude is still determined by natural numbers. To be precise, \(\mathbb{R}\) has the Archimedan property.

More precisely, \(\mathbb R\) has the Archimedean property.

Archimedan Property

Archimedan Property

There are three equivalent formulations.

  • For any positive real number \(\varepsilon\) (epsilon) and any real number \(x\), there is a natural number \(N\) so that \[N\varepsilon >x.\]

  • For any real number \(x\), there is a natural number \(N\) so that \[N > x.\]

  • For any positive real number \(\varepsilon\) (epsilon), there is a natural number \(N\) so that \[0< \tfrac{1}{N} < \varepsilon.\]

Understand the definition with the following figure.

The first statement says that for any positive real number \(\varepsilon\) and any real number \(x\), it is always possible to find a big enough natural number so that the product \(N\varepsilon\) is bigger than \(x\).

The second statement says for any real number \(x\), you can always find a big enough natural number \(N\) that is bigger than \(x\).

The third statement says that for any positive real number \(\varepsilon\), you can find a natural number \(N\) so its reciprocal is smaller than \(\varepsilon.\)

The Archimedean property of \(\mathbb R\) has some immediate consequences for infinite unions and intersections.

Archimedean Property Applied to Unions and Instructions

If \(x\) is a real number, then

  1. \(\displaystyle \bigcup_{n\in\mathbb N} \left(-\infty,x-\tfrac{1}{n}\right] = (-\infty, x)\);
  2. \(\displaystyle \bigcup_{n\in\mathbb N} \left[x+\tfrac{1}{n}, \infty\right) = (x, \infty)\);
  3. \(\displaystyle \bigcap_{n\in\mathbb N} \left(-\infty, x+\tfrac{1}{n}\right) = (-\infty, x]\);
  4. \(\displaystyle \bigcap_{n\in\mathbb N} \left(x-\tfrac{1}{n}, \infty\right) = [x, \infty)\).

Understand the consequences with the following picture.

Practice using these consequences with the next example. It help to also sketch what each \(I_n\) looks like.

Example 5

For each of these choices of sequence of intervals \((I_n)\), determine the union \(\displaystyle\bigcup_{n\in\mathbb{N}} I_n\) and the intersection \(\displaystyle\bigcap_{n\in\mathbb{N}}I_n\):

  1. \(I_n=\left[0-\frac{1}{n},1+\frac{1}{n}\right]\)
  2. \(I_n=\left[0+\frac{1}{n},1-\frac{1}{n}\right]\)
  3. \(I_n=\left(0+\frac{1}{n},1+\frac{1}{n}\right)\)

  1. Notice that \(I_1=[-1,2]\) and \(a_n=0-\frac{1}{n}\) is an increasing sequence while \(b_n=1+\frac{1}{n}\) is a decreasing sequence. So, we have that \[\bigcup_{n\in\mathbb{N}}I_n=[-1,2]\quad\text{and}\quad \bigcap_{n\in\mathbb{N}}I_n=[0,1].\]

  2. Notice that \(I_1=[1,0]=\emptyset\), \(I_2=[\tfrac{1}{2},\tfrac{1}{2}]\) and \(a_n=0+\frac{1}{n}\) is a decreasing sequence while \(b_n=1-\frac{1}{n}\) is an increasing sequence. So, we have that \[\bigcup_{n\in\mathbb{N}}I_n=(0,1)\quad\text{and}\quad \bigcap_{n\in\mathbb{N}}I_n=\emptyset.\] In the case of the intersection, it is empty because the first interval, \(I_1\) is the emptyset.

  3. Notice that \(I_1=(-1,2)\), and \(a_n=0-\frac{1}{n}\) is an increasing sequence while \(b_n=1+\frac{1}{n}\) is an decreasing sequence. So, we have that \[\bigcup_{n\in\mathbb{N}}I_n=(-1,2)\quad\text{and}\quad \bigcap_{n\in\mathbb{N}}I_n=[0,1].\]

In the next example, we will use the Archimedean property to determine what the following union and intersection equals.

Example 6

Use the Archimedean property of \(\mathbb{R}\) to express these sets as intervals:

  1. \(\displaystyle \bigcup_{n\in\mathbb{N}}\left[\frac{1}{\sqrt{n+2}},2\right]\)
  2. \(\displaystyle \bigcap_{n\in\mathbb{N}}\left[0,4+\frac{3n^2+3n+5}{n^2+1}\right]\)
  1. The right endpoint of each interval always contains \(2\), so that will be in our union. For any positive real number \(x\) and any natural number \(n\), rewrite \[\frac{1}{\sqrt{n+2}}<x\quad\text{as}\quad \frac{1}{x^2}-2<n.\] Use the Archimedean property ii of \(\mathbb{R}\) on \(\frac{1}{x^2}-2\) to obtain that there is a natural number \(N\) so that \[N>\frac{1}{x^2}-2,\] which is equivalent to \[x>\frac{1}{\sqrt{N+2}}.\] Therefore, for any natural \(n>N\), \[0<\frac{1}{\sqrt{n+2}}<\frac{1}{\sqrt{N+2}}<x.\] So for any \(x\) in \((0,2]\), there is an \(n\) so that \(x\) is in the interval \(\left[\frac{1}{\sqrt{n+2}},2\right].\) However, the number \(0\) is not in any of the intervals in the union, so it is not in the union, which means that \[\displaystyle \bigcup_{n\in\mathbb{N}}\left[\frac{1}{\sqrt{n+2}},2\right]=(0,2].\]
  2. The left endpoint of each interval always contains \(0\), so that will be in our intersection. Since \(\frac{1}{n^2+1}<\frac{1}{n^2}\), we have that \[\frac{3n^2+3n+5}{n^2+1}<\frac{3n^2+3n+5}{n^2}=3+\frac{3}{n}+\frac{5}{n^2}.\] Therefore, for any positive real number \(\varepsilon\), the inequality \[4+\frac{3n^2+3n+5}{n^2+1}<7+\varepsilon\] is valid for any natural number \(n\) as long as \(\frac{3}{n}+\frac{5}{n^2}<\varepsilon.\) Because \(\frac{1}{n^2}\leq \frac{1}{n}\) for any natural number \(n\), the inequality \(\frac{3}{n}+\frac{5}{n^2}<\varepsilon\) is valid as long as \[\varepsilon>\frac{3}{n}+\frac{5}{n^2}>\frac{3}{n}+\frac{5}{n}=\frac{8}{n}.\] By the Archimedean property, there is a naural number \(N\) so that \[N\varepsilon>8\quad\text{which is equivalent to}\quad \varepsilon>\frac{8}{N}.\] This implies that for any natural number \(n>N\), \(\frac{8}{n}<\varepsilon\). Altogether, we get that for any natural number \(n>N\), \[4+\frac{3n^2+3n+5}{n^2+1}<7+\varepsilon.\] Therefore, for any \(x>7\), there is a natural number \(n\) so that \(x>4+\frac{3n^2+3n+5}{n^2+1}.\) Since \(7\) is in each set in the intersection, this means that \[\displaystyle \bigcap_{n\in\mathbb{N}}\left[0,4+\frac{3n^2+3n+5}{n^2+1}\right]=[0,7].\]

In the previous two examples, we “computed” what the sequence of unions and sequence of intersections produced. This infinite process produced a tangible thing: an interval.

We give a name to this process in the next subsection.

But, if we move to the next subsection, we will discuss another important difference between the real numbers, \(\mathbb{R}\), and rational numbers, \(\mathbb{Q}\).

The real numbers \(\mathbb{R}\) have the nested interval property:

Nested Interval Property

For any sequence \((I_n)\) of closed and bounded, nonempty intervals in \(\mathbb{R}\) with the property that for each natural number \(n\), \(I_{n+1}\) is contained in \(I_n\), \[\bigcap_{n\in\mathbb{N}}I_n\not=\emptyset.\]

What this tells us is that the real numbers do not have holes, so the intersection is nonempty. However, the rational numbers do have holes.

To see this, take \((a_n)\) to be an increasing sequence of rational numbers so that for each \(n\), \(a_n\) is in \((\sqrt{2}-\frac{1}{n},\sqrt{2})\). Take \((b_n)\) to be a decreasing sequence of rational numbers so that for each \(n\), \(b_n\) is in \((\sqrt{2},\sqrt{2}+\frac{1}{n})\).

For each \(n\) take \(I_n\) to be \[I_n=[a_n,b_n]\bigcap \mathbb{Q}.\]

The sequence \((I_n)\) is a sequence of nested intervals in \(\mathbb{Q}\), but \(\cap I_n\) is empty because \(\sqrt{2}\) is not rational.

This is why we will work with real numbers going forward.

Sequential Limits and the Limit Laws

A sequence can be a collection of random numbers. It can also be a collection of numbers with some sort of pattern.

The terms of a sequence can get closer and closer to a single value, just like rational functions and exponential functions got closer and closer to their horizontal asymptotes.

We define this idea in the following definition.

Convergent Sequence

A sequence \((a_n)\) in \(\mathbb R\) converges to a real number \(L\) if:

  • For any positive real number \(\varepsilon\)
  • there is a natural number \(N\)
  • so that for any natural number \(n\),
  • if \(n\) is larger than \(N\), then \(|a_n - L| < \varepsilon\).

What does this rather complicated statement mean? It means that as long as \(n\) is large enough, \(a_n\) is as close as we like to \(L\).

Use the figure below to visualize this idea.

When a sequence converges to a real number, we use the following notation.

Limit of a Sequence

The number \(L\) is said to be the limit of the sequence \((a_n)\) and we denote this symbolically by \[\lim_{n\to \infty} a_n = L \quad\text{or}\quad a_n \to L\] which reads: “the limit as \(n\) tends to infinity of \(a_n\) is equal to \(L\)” or “\((a_n)\) tends to \(L\) as \(n\) tends to \(\infty\).”

Here are two examples of a sequence.

Example 7

The sequences \(\big(\frac{1}{n}\big)\) and \(\big(-\frac{1}{n}\big)\) both converge to \(0\). Meaning, \[\lim_{n\to\infty}\frac{1}{n}=0\] and \[\lim_{n\to\infty}\left(-\frac{1}{n}\right)=0.\] The sequence \(a_n=\frac{1}{n}\) is decreasing toward 0 while \(b_n=-\frac{1}{n}\) increases toward 0. Show that both sequences converge to \(0\).

  1. For any positive real number \(\varepsilon\) and any natural number \(n\) for which the inequality \[\frac{1}{n}<\varepsilon\] holds, this is equivalent to \[\frac{1}{\varepsilon}<n.\] The Archimedean property of \(\mathbb{R}\) implies that there is a natural number \(N\) so that \[N>\frac{1}{\varepsilon}\quad\text{which is equivalent to}\varepsilon>\frac{1}{N}.\] Therefore, for \(n>N\), we have that \[\frac{1}{n}<\frac{1}{N}<\varepsilon.\] Hence, for \(n>N\), \[\begin{align*} |\frac{1}{n}-0|&=|\frac{1}{n}|\\ &=\frac{1}{n}\\ &<\frac{1}{N}\\ &<\varepsilon. \end{align*}\] and \[\begin{align*} |-\frac{1}{n}-0|&=|-\frac{1}{n}|\\ &=\frac{1}{n}\\ &<\frac{1}{N}\\ &<\varepsilon. \end{align*}\] Thus, \((\frac{1}{n})\) and \((-\frac{1}{n})\) both converges to \(0\).

Informally, we can think of what we learned when we did \(y\)-axis inversion. As \(n\) gets larger, \(\frac{1}{n}\) gets smaller.

A sequence does not have to converge to zero. In the next example, we construct two sequences that converge to non-zero numbers.

Example 8

Find an example of a sequence \((a_n)\) with the following properties:

  1. \((a_n)\) is strictly increasing and converges to 5;

  2. \((a_n)\) is strictly decreasing and converges to 7.

  1. The sequence \((a_n)\) defined by \(a_n=5-\frac{1}{n}\) is strictly increasing and converges to 5.

  1. The sequence \((a_n)\) defined by \(a_n=7+\frac{1}{n}\) is strictly decreases and converges to 7.

To prove that sequences converge, it is necessary to estimate the distance between real values.

One of the most important tools to make estimate is the triangle inequality, which states that for any real numbers \(x\) and \(y\),

\[|x+y|\leq |x|+|y|\quad\text{and}\quad |x-y|\leq |x|+|y|.\]

This inequality implies another inequality, which is called the reverse triangle inequality, which states that for any real numbers \(x\) and \(y\),

\[|x-y|\geq ||x|-|y||.\]

Use these two inequalities to answer the following question

Example 9

Take \(I\) and \(I'\) to be the intervals that are, respectively, the solution sets to these inequalities: \[|x+2|\leq 3\quad\text{and}\quad |y-7|\leq 5.\]

  1. Sketch \(I\) and \(I'\)
  2. Use part a to graphically determine the minimum distance between the points in \(I\) and \(I'\), as well as the smallest interval that is guaranteed to contain the sum \(x+y\).
  3. Use the triangle inequality to determine the smallest interval that is guaranteed to contain the sum \(x+y\).
  1. The first inequality \(|x+2|\leq 3\) is the same as \(|x-(-2)|\leq 3\). Therefore, \(I\) is all real numbers \(x\) that are a distance of \(3\) from \(2\). The solution set \(I'\) to the second inequality, \(|y-7|\leq 5\), is all real numbers \(y\) that are a distance of \(5\) from \(7\). So the sketches are given below.

  1. Based on the picture, the gap between \(I\) and \(I'\) is \(1\), so the minimum distance between points in \(I\) and \(I'\) is \(1\). As for \(x+y\), overlay the pictures like this:

That is, line up \(I\) on \(I'\) where \(0\) in \(I\) lines up at \(2\) and \(12\). The pictures shows that \(I+I'=[-3,13].\)

  1. Notice that \(x+2+y-7=x+y-5\). We have that \[\begin{align*} |x+y-5|&=|x+2+y-7|\\ &\leq |x+2|+|y-7|\\ &\leq 3+5\\ &=8, \end{align*}\] so \(x+y\) is an interval of length \(16\) with center \(5\). The interval is \([5-8,5+8]=[-3,13].\) So, the triangle inequality implies that \(I+I'\subseteq [-3,13].\)

In this next example, we use the reverse triangle inequality to show the following lower bound.

Example 10

For any sequence \((a_n)\) in \(\mathbb{R}\) that converges to \(3\), show that if \(n\) is large enough, then \[|a_n-4|>\frac{1}{2}.\]

Because \(a_n\) converges to \(3\), there is a natural number \(N\) so that for all \(n>N\), \[|a_n-3|<\frac{1}{2}.\] Therefore, if \(n\) is greater than \(N\), then using the reverse triangle inequality, we have that \[\begin{align*} |a_n-4|&=|a_n-3-1|\\ &\geq ||a_n-3|-1|. \end{align*}\] Because \(|a_n-3|<\frac{1}{2}\) for all \(n\geq N\), we have that \[\begin{align*} |a_n-3|&<\frac{1}{2}\\ |a_n-3|-1&<\frac{1}{2}-1\\ |a_n-3|-1&<-\frac{1}{2}, \end{align*}\] which implies that \[||a_n-3|-1|>\frac{1}{2}.\] Thus \[\begin{align*} |a_n-4|&>\frac{1}{2}. \end{align*}\]

In this next example, we will directly use the Archimedean property show the following sequences converge. In each problem, it may be helpful to start with estimating what the distance looks like then applying the Archimedean property

Example 11

Use only the Archimedean property of \(\mathbb{R}\) to show that for any positive real number \(\varepsilon\), there is a natural number \(N\) so that if \(n\) is greater than \(N\), then \[|a_n-L|<\varepsilon,\] for the following choices of \((a_n)\) and \(L\):

  1. \(a_n=\frac{5n-2}{n+1}\) and \(L=5\)
  2. \(a_n=\sqrt{9+\frac{1}{n}}\) and \(L=3\).
  1. Notice that \[\begin{align*} |a_n-5|&=\left|\frac{5n-2}{n+1}-5\right|\\ &=\left|\frac{5n-2-5(n+1)}{n+1}\right|\\ &=\left|\frac{-7}{n+1}\right|\\ &=\frac{7}{n+1}. \end{align*}\] Therefore, for any positive real number \(\varepsilon\) and any natural number \(n\) for which \[\frac{7}{n+1}=|a_n-5|<\varepsilon,\] this is equivalent to \[n>\frac{7}{\varepsilon}-1.\] The Archimedean property of \(\mathbb{R}\) implies that there is a natural number \(N\) so that \[N>\frac{7}{\varepsilon}-1\] which is equivalent to \[\frac{7}{N+1}<\varepsilon.\] Therefore for \(n>N\) we have that \[\begin{align*}|a_n-5|&=\frac{7}{n+1}\\ &<\frac{7}{N+1}\\ &<\varepsilon. \end{align*}\]
  2. Notice that \[\begin{align*} |a_n-3|&=\left|\sqrt{9+\frac{1}{n}}-3\right|\\ &=\sqrt{9+\frac{1}{n}}-3\\ &=\left(\sqrt{9+\frac{1}{n}}\right)\cdot\left(\frac{\sqrt{9+\frac{1}{n}}+3}{\sqrt{9+\frac{1}{n}}+3}\right)\\ &=\frac{9+\frac{1}{n}-9}{\sqrt{9+\frac{1}{n}}+3}\\ &=\frac{\frac{1}{n}}{\sqrt{9+\frac{1}{n}}+3}\\ &=\frac{1}{n}\cdot\left(\frac{1}{\sqrt{9+\frac{1}{n}}+3}\right). \end{align*}\] For any natural number \(n\), \[\sqrt{9+\frac{1}{n}}+3>\sqrt{9}+3=6\] implies that \[\frac{1}{\sqrt{9+\frac{1}{n}}+3}<\frac{1}{6},\] thus \[\begin{align*} |a_n-3|&=\frac{1}{n}\cdot\left(\frac{1}{\sqrt{9+\frac{1}{n}}+3}\right)\\ &<\frac{1}{n}\cdot\frac{1}{6}. \end{align*}\] Therefore, for any positive real number \(\varepsilon\) and any natural number \(n\) for which \[\frac{1}{6n}<\varepsilon,\quad\text{which is equivalent to}\quad \frac{1}{6\varepsilon}<n,\] this implies that \[|a_n-3|<\varepsilon.\] The Archimedean property of \(\mathbb{R}\) implies that there is a natural number \(N\) so that \[N>\frac{1}{6\varepsilon}\] which is equivalent to \[\frac{1}{6N}<\varepsilon.\] Therefore for \(n>N\) we have that \[\begin{align*}|a_n-3|&<\frac{1}{6n}\\ &<\frac{1}{6N}\\ &<\varepsilon. \end{align*}\]

Any sequence can be decomposed into two parts in the following way: for any natural number \(N\) and any sequence \((a_n)\),

  • the head of \((a_n)\) is the finite sequence \((a_n)_{n\in\{1,\dots, N\}}\),
  • the tail of \((a_n)\) is the sequence \((a_{N+n})\).

The tail of a sequence is especially important for convergent sequences because for any positive real number \(\varepsilon\), there is a natural number \(N\) so that \((a_{n+N})\) is a sequence in the interval \((L-\varepsilon,L+\varepsilon).\) In otherwords, the tail of a sequence can be as close as required to the limit \(L\) for the right choice of \(N\).

Here are two basic facts about the range of a convergent sequence in \(\mathbb R\) by thinking of the tail of a convergent sequence \((a_n)\).

Theorem on Convergent Sequences

  1. For any convergent sequence \((a_n)\), \(a(\mathbb N)\) is bounded. Meaning there are real numbers \(M\) and \(N\) so that \(N \leq a_n\leq M\) for all \(n\).

  2. For any sequence \((b_n)\) that converges to a nonzero real limit, if for all \(n\), \(b_n\) is nonzero, then there is positive real number \(\varepsilon\) so that \[|b_n| > \varepsilon.\]

  3. If the limit of a convergent sequence \(c_n\) is positive (negative), then there is a natural number \(N\) so that for all \(n\) greater than \(N\), \(c_n\) is positive (negative).

Do you see why these statements are true? Think of the tail of the sequence!

They are true because if a sequence converges to a limit \(L\), then for any positive real number \(\varepsilon\), only finitely many terms of the sequence are outside the interval \((L-\varepsilon, L +\varepsilon)\), and so the limit “controls” all but finitely many terms of a convergent sequence.

To help us understanding more properties of a sequence, we first focus on a specific kind of sequence.

Null Sequence

The sequence \((a_n)\) is a null sequence if \((a_n)\) is convergent to \(0\).

It is convenient to prove statements about sequences by proving them first for null sequences. Think of null sequences as describing errors in finite approximation procedures. It is very useful to study such procedures by carefully studying the errors.

The equalities \[|a_n-0| = ||a_n|-0|,\quad |a_n-0| = |(a_n+L)-L| \quad \text{and} \quad |a_n-L| = |(a_n-L)-0|\] respectively imply the following:

Theorem on Null Sequences

  1. \((a_n)\) is null sequence if and only if \((|a_n|)\) is a null sequence;
  2. \((a_n)\) is a null sequence if and only if \((a_n +L)\) is convergent to \(L\);
  3. \((a_n)\) is convergent to \(L\) if and only if \((a_n -L)\) is a null sequence

The first statement tells us a sequence is null if and only if the absolute value of its terms is also a null sequence. The second statement tells us a sequence is null if and only if adding a constant to its terms converges to that constant. Finally, the third statement tell us that a sequence is convergent to \(L\) if and only if the difference of the sequence and its limit \(L\) is a null sequence.

We also have the following theorem as well.

Limit Laws for Null Sequences

  1. (Sandwich Theorem for null sequences) For any sequences \((a_n)\), \((b_n)\), and \((c_n)\) in \(\mathbb R\), if \((a_n)\) and \((c_n)\) are null sequences and for all \(n\), \[a_n \leq b_n \leq c_n,\] then \((b_n)\) is a null sequence.

  2. (Sum law for null sequences) For any null sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), \((a_n+b_n)\) is a null sequence.

  3. (Product lmit law for null sequences) For any null sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), \((a_nb_n)\) is a null sequence.

  4. (Quotient law for null sequences) For any sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), if \((a_n)\) is a null sequence, \((b_n)\) is convergent to a nonzero limit \(L\), and for all \(n\), \(b_n\) is nonzero, then \(\big(\frac{a_n}{b_n}\big)\) is a null sequence.

Convergence theorems involving null sequences have a wide range of application, even though null sequences are very special kinds of sequences. As we will see, all convergence theorems for sequences that we will discuss follow from the convergence theorems that we have already presented.

The next examples shows the usefulness of explicitly writing error terms as a sequence.

Example 12

  1. Take \(a_n\) to be the sequence that is given by \[a_n = 2^{\frac{1}{n}}.\] Use the fact that there is a null sequence \((\varepsilon_n)\) so that for each \(n\), \[2^\frac{1}{n}=1+\varepsilon_n,\] to calculate \(\displaystyle\lim_{n\to \infty} a_n\).

  2. Calculate \(\displaystyle\lim_{n\to \infty} \sqrt{9+\frac{1}{n}}\). Carefully justify your reasoning.

  1. For each natural number \(n\), \(2^{\frac{1}{n}}\) is greater than \(1\). Therefore, there is a positive real number \(\varepsilon_n\) so that \[2^\frac{1}{n}=1+\varepsilon_n,\] Take the \(n^{\text{th}}\) power of both sides to obtain the equality \[2=(1+\varepsilon_n)^n.\] Positivity of \(\varepsilon_n\) implies that \((1+\varepsilon_n)^n>1+n\varepsilon_n,\) and so \[2>1+n\varepsilon_n.\] Rewrite the inequality to get \[0<\varepsilon_n<\frac{1}{n}.\] Since the error term is bounded by two null sequences, conclude by the Sandwich Theorem for nul squences that \[\displaystyle\lim_{n\to\infty}\varepsilon_n=0.\] Thus, \[\displaystyle \lim_{n\to\infty}2^{\frac{1}{n}}=\displaystyle \lim_{n\to\infty}(1+\varepsilon_n)=1+0=1.\]

  2. For each \(n \in\mathbb{N}\), there is a positive number \(\varepsilon_n\) (error term) so that \(\sqrt{9+\frac{1}{n}}=3+\varepsilon_n\). This means that \[\left(\sqrt{9+\frac{1}{n}}\right)^2=(3+\varepsilon_n)^2\geq 3^2+2\cdot 3\varepsilon_n\] or \[9+\frac{1}{n}\geq 9+6\varepsilon_n.\] Solve for \(\varepsilon\) to get \[0<\varepsilon_n<\frac{1}{6n}.\] Since the error term is bounded by two null sequences, conclude by the Limit Laws for Null Sequences that \[\displaystyle\lim_{n\to\infty}\varepsilon_n=0.\] Thus, \[\displaystyle \lim_{n\to\infty}\sqrt{9+\frac{1}{n}}=\displaystyle \lim_{n\to\infty}(3+\varepsilon_n)=3+0=3.\]

General formulas, known as the limit laws, are very useful for calculating the limits of sequences.

Limit Laws for Convergent Sequences

For any sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\) that converge respectively to limits \(L\) and \(M\), \((a_n+b_n)\) and \((a_nb_n)\) are both convergent and

  1. (limit law for sums) \(\displaystyle\lim_{n\to\infty} (a_n+b_n) = L+M\);

  2. (limit law for products) \(\displaystyle\lim_{n\to\infty} a_nb_n=LM\).

  3. (limit law for quotients) In addition, if for all \(n\), \(b_n\) is nonzero and \(M\) is not equal to \(0\), then \(\left(\frac{a_n}{b_n}\right)\) is convergent and \(\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{L}{M}\).

These results follow by rewriting the sequences like so.

  • \(a_n+b_n=a_n+b_n+L-L+M-M=(a_n-L)+(b_n-M)+L+M\),
  • \(a_nb_n =a_nb_n+Lb_n-b_nL+LM-LM=b_n(a_n-L)+L(b_n-M)+LM\)
  • \(\frac{a_n}{b_n}=\left(a_n\right)\cdot \left(\frac{1}{b_n}\right)=a_n\cdot \left(\frac{1}{b_n}-\frac{1}{M}+\frac{1}{M}\right)=a_n\cdot \left(\frac{1}{Mb_n}(b_n-M)+\frac{1}{M}\right)\)

We rewrite the above terms so that we introduce null sequences and can use the properties of null sequences in the previous theorem to obtain the results.

Example 13

Take \((a_n)\), \((b_n)\), and \((c_n)\) to be sequences with \[ \lim_{n\to\infty}a_n = 2, \quad \lim_{n\to\infty}b_n = 3, \quad \text{and} \quad \lim_{n\to\infty}c_n = 7.\] Use the limit laws to compute the following:

  1. \(\displaystyle \lim_{n\to\infty}a_nb_nc_n\)

  2. \(\displaystyle \lim_{n\to\infty}a_n^3-5b_n\)

  3. \(\displaystyle \lim_{n\to\infty}\tfrac{a_n+b_n}{c_n}\)

  1. Use limit law for products to get \[\begin{align*} \lim_{n\to\infty}a_nb_nc_n&=\lim_{n\to\infty}\left(a_nb_n\right)c_n\\ &=\lim_{n\to\infty}\left(a_nb_n\right)\cdot \lim_{n\to\infty}c_n\\ &=\lim_{n\to\infty}a_n\cdot \lim_{n\to\infty} b_n\cdot \lim_{n\to\infty}c_n\\ &=2\cdot 3\cdot 7\\ &=42 \end{align*}\]

  2. Use limit law for products and sums to get \[\begin{align*} \lim_{n\to\infty}\left(a_n^3-5b_n\right)&=\lim_{n\to\infty}\left(a_n^3+(-5b_n)\right)\\ &=\lim_{n\to\infty}(a_n^3)+\lim_{n\to\infty}(-5\cdot b_n)\\ &=\lim_{n\to\infty}(a_n\cdot a_n\cdot a_n)+\lim_{n\to\infty}(-5\cdot b_n)\\ &=\lim_{n\to\infty}a_n\cdot \lim_{n\to\infty}a_n\cdot \lim_{n\to\infty}a_n+\lim_{n\to\infty}(-5)\cdot\lim_{n\to\infty} b_n\\ &=2\cdot 2\cdot 2+(-5)\cdot 3 \\ &=8-15\\ &=-7 \end{align*}\]

  3. Use limit law for quotient and sums to get \[\begin{align*} \lim_{n\to\infty}\tfrac{a_n+b_n}{c_n}&=\lim_{n\to\infty}\tfrac{(a_n+b_n)}{c_n}\\ &=\tfrac{\displaystyle\lim_{n\to\infty}(a_n+b_n)}{\displaystyle\lim_{n\to\infty}c_n}\\ &=\tfrac{\displaystyle\lim_{n\to\infty}a_n+\displaystyle\lim_{n\to\infty}b_n}{\displaystyle\lim_{n\to\infty}c_n}\\ &=\tfrac{2+3}{7}\\ &=\tfrac{5}{7} \end{align*}\]

In the next examples, we cannot use the limit laws immediately because the numerator and denominator of the sequence do not converge separately. However, every sequence does converge and this can be seen by rewriting the quotient in the right way.

Example 14

Use the limit laws to determine the following limits:

  1. \(\displaystyle \lim_{n\to\infty}\tfrac{5n+1}{n+3}\)
  2. \(\displaystyle \lim_{n\to\infty}\tfrac{n^2+7n+2}{n^2+2n-1}\)
  3. \(\displaystyle \lim_{n\to\infty}\tfrac{n-4}{n^2+3n-2}\)

  1. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n}}{\frac{1}{n}}\): \[\begin{align*} \frac{5n+1}{n+3}&=\frac{5n+1}{n+3}\cdot \tfrac{\frac{1}{n}}{\frac{1}{n}}\\ &=\frac{(5n+1)\cdot \frac{1}{n}}{(n+3)\cdot \frac{1}{n}}\\ &=\frac{5n\cdot \frac{1}{n}+1\cdot \frac{1}{n}}{n\cdot\frac{1}{n}+3\cdot\frac{1}{n}}\\ &=\frac{5+\frac{1}{n}}{1+3\cdot\frac{1}{n}}.\\ \end{align*}\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\frac{5n+1}{n+3}&=\lim_{n\to\infty}\frac{5+\frac{1}{n}}{1+3\cdot\frac{1}{n}}\\ &=\frac{\displaystyle\lim_{n\to\infty}5+\displaystyle\lim_{n\to\infty}\frac{1}{n}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}3\cdot\displaystyle\lim_{n\to\infty}\frac{1}{n}}\\ &=\frac{5+0}{1+3\cdot0}\\ &=\frac{5}{1}\\ &=5. \end{align*}\]

  2. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\): \[\begin{align*} \frac{n^2+7n+2}{n^2+2n-1}&=\frac{n^2+7n+2}{n^2+2n-1}\cdot \tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\\ &=\frac{(n^2+7n+2)\cdot \frac{1}{n^2}}{(n^2+2n-1)\cdot \frac{1}{n^2}}\\ &=\frac{n^2\cdot \frac{1}{n^2}+7n\cdot \frac{1}{n^2}+2\cdot\frac{1}{n^2}}{n^2\cdot\frac{1}{n^2}+2n\cdot\frac{1}{n^2}-1\cdot\frac{1}{n^2}}\\ &=\frac{1+\frac{7}{n}+\frac{2}{n^2}}{1+\frac{2}{n}-\frac{1}{n^2}}\\ \end{align*}.\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\tfrac{n^2+7n+2}{n^2+2n-1}&=\lim_{n\to\infty}\frac{1+\frac{7}{n}+\frac{2}{n^2}}{1+\frac{2}{n}-\frac{1}{n^2}}\\ &=\frac{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{7}{n}+\displaystyle\lim_{n\to\infty}\frac{2}{n^2}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{2}{n}-\displaystyle\lim_{n\to\infty}\frac{1}{n^2}}\\ &=\frac{1+0+0}{1+0-0}\\ &=\frac{1}{1}\\ &=1 \end{align*}.\]

  3. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\): \[\begin{align*} \frac{n-4}{n^2+3n-2}&=\frac{n-4}{n^2+3n-2}\cdot \tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\\ &=\frac{(n-4)\cdot \frac{1}{n^2}}{(n^2+3n-2)\cdot \frac{1}{n^2}}\\ &=\frac{n\cdot \frac{1}{n^2}-4\cdot \frac{1}{n^2}}{n^2\cdot\frac{1}{n^2}+3n\cdot\frac{1}{n^2}-2\cdot\frac{1}{n^2}}\\ &=\frac{\frac{1}{n}-\frac{4}{n^2}}{1+\frac{3}{n}-\frac{2}{n^2}}\\ \end{align*}.\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\tfrac{n-4}{n^2+3n-2}&=\lim_{n\to\infty}\frac{\frac{1}{n}-\frac{4}{n^2}}{1+\frac{3}{n}-\frac{2}{n^2}}\\ &=\frac{\displaystyle\lim_{n\to\infty}\frac{1}{n}-\displaystyle\lim_{n\to\infty}\frac{4}{n^2}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{3}{n}-\displaystyle\lim_{n\to\infty}\frac{2}{n^2}}\\ &=\frac{0-0}{1+0-0}\\ &=\frac{0}{1}\\ &=0 \end{align*}.\]

In this next example, we use the limit laws together with sum rewriting to compute the following limits.

Example 15

Take \(x\) to be a real number. Use the limit laws to evaluate \(\displaystyle\lim_{n\to\infty}a_n\) for each of these choices of sequence \((a_n)\):

  1. \(a_n=\sqrt{x^2+\frac{1}{n}}\)
  2. \(a_n=n\left(\sqrt{x^2+\frac{1}{n^2}}-|x|\right)\), for \(x\) not equal to \(0\).
  1. Since \(\frac{1}{n}\) is always positive for each natural number and the \({\rm{pow}}_\frac{1}{2}\) is increasing, for each \(n\) there is a positive real number \(\varepsilon_n\) so that \[\sqrt{x^2+\frac{1}{n}}=\sqrt{x^2}+\varepsilon_n=|x|+\varepsilon_n.\] Square both sides of the equality to obtain \[x^2+\frac{1}{n}=\left(|x|+\varepsilon_n\right)^2=x^2+2|x|\varepsilon_n+\varepsilon_n^2.\] Since \(2|x|\varepsilon_n\) is positive, \[x^2+\frac{1}{n}=x^2+2|x|\varepsilon_n+\varepsilon_n^2>x^2+\varepsilon_n^2,\] which implies that \[x^2+\frac{1}{n}>x^2+\varepsilon_n^2\quad\text{or}\quad \frac{1}{n}>\varepsilon_n^2.\] This is equivalent to \[0<\varepsilon_n<\frac{1}{\sqrt{n}}.\] The sandwich theorem implies that \((\varepsilon_n)\) is a null sequence, which implies that \[\begin{align*} \lim_{n\to\infty}a_n&=\lim_{n\to\infty}\left(|x|+\varepsilon_n\right)\\ &=|x|. \end{align*}\]
  2. Rewrite the product like this:

\[\begin{align*} n\left(\sqrt{x^2+\frac{1}{n}}-|x|\right)&=n\left(\sqrt{x^2+\frac{1}{n}}-|x|\right)\cdot \frac{\sqrt{x^2+\frac{1}{n}}+|x|}{\sqrt{x^2+\frac{1}{n}}+|x|}\\ &=n\frac{x^2+\frac{1}{n}-x^2}{\sqrt{x^2+\frac{1}{n}}+|x|}\\ &=n\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+|x|}\\ &=\frac{1}{\sqrt{x^2+\frac{1}{n}}+|x|}. \end{align*}\] Therefore, by the quotient limit law \[\begin{align*} \lim_{n\to\infty}n\left(\sqrt{x^2+\frac{1}{n}}-|x|\right)&=\lim_{n\to\infty}\frac{1}{\sqrt{x^2+\frac{1}{n}}+|x|}\\ &=\frac{\displaystyle\lim_{n\to\infty} 1}{\displaystyle\lim_{n\to\infty}\left(\sqrt{x^2+\frac{1}{n}}+|x|\right)}\\ &=\frac{1}{|x|+|x|}\\ &=\frac{1}{2|x|}. \end{align*}\]

In this example, we will show that if we have enough information, we can show a sequence converges. Pay close attention to how we use the limit laws. Notice that we do not use the limit law for products because we don’t know that \(b_n\) converges.

Example 16

Take \((a_n)\) and \((b_n)\) to be sequences with \[\lim_{n\to\infty}a_n=3\quad \text{and} \quad \lim_{n\to\infty}a_nb_n=18.\] Carefully justify that \((b_n)\) is convergent and calculate its limit.

As long as \(a_n\not=0\), the identity \(b_n=\frac{a_nb_n}{a_n}\) holds. By Theorem on Convergent Sequences, there is an \(N\) so that \(a_n\not=0\) for \(n\geq N\). Use the limit laws to conclude that \[\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{a_nb_n}{a_n}=\frac{\displaystyle\lim_{n\to\infty}a_nb_n}{\displaystyle\lim_{n\to\infty}a_n}=\frac{18}{3}=6.\]

Now we study a special kind of sequences that are defined by their previous terms.

Recursive Sequences

A sequence \((a_n)\) is defined recursively if for each \(n\), \(a_n\) is given by some formula that involves \(n\) and the previous terms in the sequence.

As an example, for all \(n\) in \(\mathbb{N}_0=\{0,1,2,\dots\}\), define \((a_n)\) by

\[\begin{cases} a_0=1\\ a_{n+1}=(n+1)a_n. \end{cases}\]

The first five terms of the sequence of \(a_n\) are

  • \(a_0=1\)
  • \(a_1=1a_0=1.\)
  • \(a_2=2a_1=2\cdot 1\)
  • \(a_3=3a_2=3\cdot 2\cdot 1\)
  • \(a_4=4a_3=4\cdot 3\cdot 2\cdot 1\). Thus \[a_n=n!,\] for all \(n\) in \(\mathbb{N}_0\).

Practice with this next example

Example 17

These recursively defined sequences have only small differences in values at the first place:

  1. \(a_1=1\), \(a_{n+1}=a_n^2\)
  2. \(a_1=\frac{1}{2}\), \(a_{n+1}=a_n^2\)
  3. \(a_1=2\), \(a_{n+1}=a_n^2\)
  1. The first couple of terms are \(a_1=1\), \(a_2=a_1^2=1\), \(a_3=a_2^2=1\) and \(a_4=a_3^2=1.\) So \(a_n=1\) for all \(n\). Thus \((a_n)\) is the constant sequence that converges to \(1\).
  2. The first couple of terms are \(a_1=\frac{1}{2}\), \(a_2=a_1^2=\frac{1}{4}\), \(a_3=a_2^2=\frac{1}{16}\) and \(a_4=a_3^2=\frac{1}{256}.\) So \(a_n=\frac{1}{2^{2n}}\) for all \(n\). Thus \((a_n)\) is a positive null sequence.
  3. The first couple of terms are \(a_1=2\), \(a_2=a_1^2=4\), \(a_3=a_2^2=16\) and \(a_4=a_3^2=256.\) So \(a_n=2^{2n}\) for all \(n\). Thus \((a_n)\) is a positive, increasing and unbounded sequence.

In order to determine what a sequence converges to, we need to develope more theory. In particular, if we know a sequence is convergent, we may be able to compute the limit directly.

However, assuming incorrectly that a sequence is convergent can lead to incorrect conclusions.

Example 18

Take \((a_n)\) to be the recursively defined sequence that is given by

\[\begin{cases} a_1=2\\ a_{n+1}=a_n^2, \end{cases}\]

Assume (incorrectly) that \((a_n)\) is convergent and use the limit laws to determine the limit of the sequence. How can you detect that there is an error in the assumption?

If \((a_n)\) were convergent to a limit \(a_0\), then \[\lim_{n\to\infty}a_{n}=a_0.\] However, because \(a_{n+1}=a_n^2\) for all \(n\geq 2\), it must also be true that \[\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}(a_n)^2=a_0^2.\] Therefore, \[a_0=\lim_{n\to\infty}a_{n+1}=a_0^2\] implies that \[a_0^2-a_0=0\quad\text{which is equivalent to}a_0(a_0-1)=0.\] This implies that \(a_0=0\) or \(a_0=1\). However, none of these can be the limit of \((a_n)\) because \(a_n\) is an increasing sequence and \(a_1=2\).

An important observation in the above example is how the sequence was increasing and unbounded, so it could not have a limit.

One criteria that enables us to conclude a sequence is convergent is the monotone convergence theorem.

Monotone Convergence Theorem

Monotone Convergence Theorem Any monotone and bounded sequence \((a_n)\) in \(\mathbb{R}\) is convergent.

This theorem follows from the Archimedean property of \(\mathbb{R}\) and the nested interval property.

Use this theorem to show the following recursive sequence is convergent and conclude what the limit is.

Example 19

Assume that the sequence given below is bounded. Use the limit laws together with the monotone convergence theorem to determine the limi of the sequence \((a_n)\), where \[\begin{cases}a_1=0\\a_{n+1}=\sqrt{2+a_n}.\end{cases}\]

We can show by induction that the sequence \(a_n\) is increasing. The base case holds because \(a_1=0\) and \(a_2=\sqrt{2}\) imply that \(a_1<a_2.\) Assume that for \(k>1\) in \(\mathbb{N}\), we have that \(a_k<a_{k+1}\). We now need to show \(a_{k+1}<a_{k+2}\). The inequality \[a_k<a_{k+1}\] is equivalent to \[2+a_k<2+a_{k+1}\quad\text{which implies that}\quad\sqrt{2+a_k}<\sqrt{2+a_{k+1}}.\]

However \(a_{k+1}=\sqrt{2+a_k}\) and \(a_{k+2}=\sqrt{2+a_{k+1}}\), so \[\sqrt{2+a_k}<\sqrt{2+a_{k+1}}\quad\text{is equivalent to}\quad a_{k+1}<a_{k+2}.\] Thus by induction, \((a_n)\) is increasing.

We can also prove by induction that \(0\leq a_n<2\) for all \(n\). The base case holds because \(a_1=0\). Assume that for some natural number \(k>1\) that \[0\leq a_k<2.\] Notice that \(a_{k+1}=\sqrt{2+a_k}<\sqrt{2+2}=2\), hence \[0<a_{k+1}<2.\] Thus, \((a_n)\) is a bounded monotone sequence, so by the monotone convergence theorem, \((a_n)\) is convergent to some limit \(a_0\). Because \((a_n)\) is convergent to \(a_0\), we have that

\[a_0=\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{2+a_n}=\sqrt{2+a_0}.\] Square both sides to obtain that \[a_0^2=2+a_0\] and so \[a_0^2-a_0-2=0.\] The solutions this equation are \(a_0=-1\) or \(2\). However, because the sequence is positive, only \(a_0=2\) makes sense. So \((a_n)\) must converge to \(2\). Why does this make sense? Recall that by the half angle formula that

  • \(\cos(\frac{\pi}{2})=0\)
  • \(\cos(\frac{\pi}{4})=\sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}{2}}=\frac{\sqrt{2}}{2}\)
  • \(\cos(\frac{\pi}{8})=\sqrt{\frac{1+\cos\left(\frac{\pi}{4}\right)}{2}}=\frac{\sqrt{2+\sqrt{2}}}{2}\)
  • \(\cos(\frac{\pi}{16})=\sqrt{\frac{1+\cos\left(\frac{\pi}{8}\right)}{2}}=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\) and so \[\cos\left(\frac{\pi}{2^{n+1}\right)=\frac{\sqrt{2+\sqrt{2+\dots+\sqrt{2}}}}{2}\] where the inside has \(n\) nested roots. Thus \[\cos\left(\frac{\pi}{2^{n+1}\right)=\frac{a_n}{2}.\] Intiution suggest that the left side should converge to \(\cos(0)=1\), which is the case.

In this example, we will again use the monotone convergence theorem to show this sequence converges.

Example 20

Take \(r\) to be in \((0,1)\). Assume that the sequence below is bounded. Use the limit laws together with the monotone convergence theorem to determine the limit of the sequence \((a_n)\), where \[ \begin{cases} a_1=1 a_{n+1}=1+ra_n. \end{cases} \]

Since \(r\) is in \((0,1)\), the sequence is increasing. We can show this by induction, but the idea can be seen by looking at the first few terms:

  • \(a_1=1\)
  • \(a_2=1+r\) or \(a_1+r\)
  • \(a_3=1+r(1+r)=1+r+r^2\) or \(a_2+r^2\)
  • \(a_4=1+r(1+r+r^2)=1+r+r^2+r^3\) or \(a_3+r^3\)

We can see that \[a_{n+1}=a_n+r^n,\] where \(a_n\) and \(r^n\) are positive. Therefore, the sequence is increasing. Use the monotone convergence theorem to conclude that \((a_n)\) is convergent to a limit \(a_0\). This means that \[\lim_{n\to\infty}a_n=a_0.\] However, \(a_{n+1}=1+ra_n\), so it is also true that \[\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\left(1+ra_n\right)=1+ra_0.\] This means that \[a_0=1+ra_0\] which is equivalent to \[a_0-ra_0=1\quad\text{or}\quad a_0=\frac{1}{1-r}.\] The sequence thus converges to \(\frac{1}{1-r}.\)

The squeeze theorem (often also called the sandwich theorem) is a valuable tool for determining the limit of a sequence.

To apply this theorem to determine the limit of a sequence \((b_n)\), the strategy is to find two sequences \((a_n)\) and \((c_n)\) so that for all \(n\), \[a_n \leq b_n\leq c_n\] and \((a_n)\) and \((c_n)\) converge to the same limit.

The sequence \((b_n)\) will then be convergent and will converge to the same limit as its bounding sequences.

Squeeze Theorem

The squeeze theorem states that if there are sequences \((a_n)\), \((b_n)\) and \((c_n)\) so that for all \(n\), \[a_n \leq b_n\leq c_n\] and \((a_n)\) and \((c_n)\) converge to the same limit, then \((b_n)\) will then be convergent and will converge to the same limit as its bounding sequences.

Finding such bounding sequences then becomes the main obstacle in applications. Use the squeeze theorem for all these problems.

Example 21

Take \(b_n\) to be the sequence that is given by \[b_n = (3^n +2^n)^{\frac{1}{n}}.\] Calculate \(\displaystyle\lim_{n\to \infty} b_n\). Carefully justify your reasoning.

Find two sequences \(a_n\) and \(c_n\) so that \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}c_n.\)

  • find upper bound: Notice that \(2^n\leq 3^n\) for all \(n\in\mathbb{N}\). So \[3^n+2^n\leq 3^n+3^n=2\cdot 3^n.\] Since \(\rm{pow}_\frac{1}{n}\) is an increasing function when \(n\in\mathbb{N}\), \[3^n+2^n\leq 2\cdot 3^n\] implies that \[\begin{align*}\left(3^n+2^n\right)^{1/n}&\leq \left(2\cdot 3^{n}\right)^{1/n}\\&\leq 2^{1/n}\cdot3\end{align*}.\] Thus \(\left(3^n+2^n\right)^{1/n}\leq 3\cdot 2^{1/n}.\)

  • find lower bound: Now notice that \(2^n\) and \(3^n\) are always positive. So \[3^n+2^n\geq 3^n.\] Since \(\rm{pow}_\frac{1}{n}\) is an increasing function when \(n\in\mathbb{N}\), \[3^n+2^n\geq 3^n\] implies that \[\begin{align*}\left(3^n+2^n\right)^{1/n}&\geq \left(3^n\right)^{1/n}\\&\geq 3\end{align*}.\] Thus \(\left(3^n+2^n\right)^{1/n}\geq 3.\)

Take \(a_n=3\) and \(c_n= 3\cdot 2^{n}\). The conclusion from the work before was \(a_n\leq b_n\leq c_n\) for all \(n\in\mathbb{N}\).

  • show upper and lower bound have same limit: \[\displaystyle \lim_{n\to\infty}a_n=\displaystyle \lim_{n\to\infty}3=3\] and \[\displaystyle \lim_{n\to\infty}c_n=\displaystyle \lim_{n\to\infty}\left(3\cdot 2^{1/n}\right)=\displaystyle \lim_{n\to\infty}3\cdot \displaystyle \lim_{n\to\infty}2^{1/n}=3\cdot 1=3.\]

Since \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}c_n=3\), apply the squeeze theorem to conclude that \(\displaystyle\lim_{n\to\infty}(3^n +2^n)^{\frac{1}{n}}=3\).

In the next example, we show a sequence converges using the squeeze theorem. Pay attention to how we find the upper bounds and lower bounds.

Example 22

Take \(a_n\) to be the sequence that is given by \[a_n = \tfrac{3n^2 + n\sin(n)}{n^2+1}.\] Calculate \(\displaystyle\lim_{n\to \infty} a_n\). Carefully justify your reasoning.

Find two sequences \(b_n\) and \(c_n\) so that \(b_n\leq a_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}b_n=\displaystyle\lim_{n\to\infty}c_n.\)

  • find upper bound: Recall that the sine function has a range of \([-1,1]\). This means that \(-1\leq \sin(n)\leq 1\) for all \(n\in\mathbb{N}\). Since \(3n^2\) and \(n\) are positive numbers, \[3n^2+n\sin(n)\leq 3n^2+n\cdot 1= 3n^2+n.\] Since \(n^2+1\) is always positive, \[3n^2+n\sin(n)\leq 3n^2+n\] implies that \[\frac{3n^2+n\sin(n)}{n^2+1}\leq \frac{3n^2+n}{n^2+1}.\]

  • find lower bound: Recall that the sine function has a range of \([-1,1]\). This means that \(-1\leq \sin(n)\leq 1\) for all \(n\in\mathbb{N}\). Since \(3n^2\) and \(n\) are positive numbers, \[3n^2+n\sin(n)\geq 3n^2+n\cdot(-1)= 3n^2-n.\] Since \(n^2+1\) is always positive, \[3n^2+n\sin(n)\geq 3n^2-n\] implies that \[\frac{3n^2+n\sin(n)}{n^2+1}\geq \frac{3n^2-n}{n^2+1}.\]

Let \(b_n=\frac{3n^2-n}{n^2+1}\) and \(c_n= \frac{3n^2+n}{n^2+1}\). The conclusion from the work before wast \(b_n\leq a_n\leq c_n\) for all \(n\in\mathbb{N}\).

  • show upper and lower bound have same limit: \(\displaystyle \lim_{n\to\infty}b_n=\displaystyle \lim_{n\to\infty}\frac{3n^2-n}{n^2+1}=4\) and \(\displaystyle \lim_{n\to\infty}c_n=\displaystyle \lim_{n\to\infty}\frac{3n^2-n}{n^2+1}=4\) (Work this out by following a similar idea in Example 10)

Since \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}b_n=\displaystyle\lim_{n\to\infty}c_n=4\), apply the squeeze theorem to conclude that \(\displaystyle\lim_{n\to\infty}\tfrac{3n^2 + n\sin(n)}{n^2+1}=4\).

Not every sequence converges. In fact, we have already seen sequences that don’t converge. For example, \(a_n=n\).

We define how certain sequences behave when they don’t converge.

Diverges to Infinity

A sequence \((a_n)\) is said to diverge to \(\infty\) (read: “infinity”) if for any natural number \(M\), there is a natural number \(N\) so that for all \(n\) in \(\mathbb N\), \[n> N \quad \text{implies that} \quad a_n > M.\]

To indicate that \((a_n)\) diverges to \(\infty\), write \[\lim_{n\to \infty} a_n = \infty \quad \text{or}\quad a_n \to \infty;\]

This means that \(a_n\) is as large as we like as long as \(n\) is large enough.

A sequence can also diverge to negative infinity.

Diverges to Negative Infinity

A sequence \((a_n)\) is said to diverge to \(-\infty\) (read: “negative infinity”) if for any natural number \(M\), there is a natural number \(N\) so that for all \(n\) in \(\mathbb N\), \[n> N \quad \text{implies that} \quad a_n < -M.\]

To indicate that \((a_n)\) diverges to \(-\infty\), write \[\lim_{n\to \infty} a_n = -\infty \quad \text{or}\quad a_n \to -\infty.\]

Use the figure below to visualize this type of divergence.

People will often say that \((a_n)\) tends to \(\infty\) or that \((a_n)\) tends to \(-\infty\) to mean that it diverges to \(\infty\) or \(-\infty\), respectively. The notation \(a_n\to\infty\) and \(a_n\to-\infty\) is sometimes used as well.

In the next example, practice identifying what kind of divergence each sequence has.

Example 23

Determine whether the sequence diverges to either infinity or negative infinity:

  1. \(a_n = 2^n\);
  2. \(a_n = n^{1/3}\);
  3. \(a_n = -n^3\).

  1. For any real number \(K\), there is a natural number so that \(N>K\). Because \(\star<2^\star\) holds for any natural number \(\star\), then for any \(n>N\), we have that \[K<N<n<2^n\] and so \[\displaystyle \lim_{n\to\infty}2^n=\infty.\]

  1. For any real number \(K\), there is a natural number \(N\) so that \(N>K\). Since \((N^3)^\frac{1}{3}=N\) holds, then for any \(n>N^3\), we have that \[K<N=(N^3)^\frac{1}{3}<n^\frac{1}{3}.\] Therefore, \[\displaystyle \lim_{n\to\infty}n^{1/3}=\infty.\]

  1. For any real number \(K\), there is a natural number \(N\) so that \(N>K\) or \(-N<-K\). Since \(\star^3>\star\) holds for all natural numbers \(\star\), then if \(n>N\), we have that \[-K>-N>-N^3>-n^3.\] Therefore, \(\displaystyle \lim_{n\to\infty}-n^3=-\infty\)

Sequences may diverge, but not tend toward infinity or negative infinity. In which case, we say it is divergent.

Example 24

For each of these choices of sequences \((a_n)\), determine whether the sequence is divergent or diverges to \(\infty\) or \(-\infty\)

  1. \(a_n=n^2\)
  2. \(a_n=\left(-1\right)^n\)
  3. \(\begin{cases}a_1=1\\a_{n+1}=a_n+\frac{1}{a_n}\end{cases}\)
  1. This sequence diverges to \(\infty\)
  2. This sequence can be rewritten like this: \[a_n=\begin{cases}1&\text{if }n\text{ is even}\\ -1&\text{if }n\text{ is odd.}\end{cases}\] This sequence is therefore divergent and it alternates between \(-1\) and \(1\).
  3. Because \(a_1=1\), \(a_1\) is positive. Moreover, for any \(n\), if \(a_n\) is positive then \(a_{n+1}\) is positive because it is the sum of the positive numbers \(a_n\) and \(\frac{1}{a_n}\). Therefore, by induction, \((a_n)\) is a positive sequence. For any natural number \(n\), \[a_{n+1}-a_n=\frac{1}{a_n},\] which is positive. Therefore, the sequence is increasing. However, it is not bounded. If the sequence was, then the Monotone Convergence Theorem implies that \((a_n)\) is convergent to some limit \(a_0\). Then by the limit laws \[a_0=\lim_{n\to\infty}a_{n+1}=a_0+\frac{1}{a_0},\] which is equivalent to \[0=\frac{1}{a_0}.\] However, there is no real number so that this is true. Thus \((a_n)\) is not bounded. Therefore, because \((a_n)\) is increasing and unbounded, it diverges to \(\infty.\)

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